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Question Number 117403 by bemath last updated on 11/Oct/20

$$\underset{0}{\stackrel{1}{\int }}{\left(\mathrm{arc}\tan \mathrm{x}\right)}^2\mathrm{dx}=?$$

Commented by MJS_new last updated on 11/Oct/20

$$\mathrm{use}\arctan x=\frac{\ln (1+\mathrm{i}x)-\ln (1-\mathrm{i}x)}{2\mathrm{i}}\Rightarrow$$$$-\frac{1}{4}\underset{0}{\stackrel{1}{\int }}({\ln }^2(1+\mathrm{i}x)-2\ln (1+\mathrm{i}x)\ln (1-\mathrm{i}x)+{\ln }^2(1-\mathrm{i}x))dx$$$$\mathrm{and}\mathrm{these}\mathrm{are}\mathrm{solveable}\left(\mathrm{by}\mathrm{parts}\right)$$

Answered by mindispower last updated on 11/Oct/20

$$={\left[{xarctan}^2\left(x\right)\right]}_0^1-{\int }_0^1\frac{2x}{1+x^2}arctan\left(x\right)dx$$$$=\frac{{\pi }^2}{16}-I$$$$letarctan\left(x\right)=t$$$$I=-{\int }_0^{\frac{\pi }{4}}\frac{2tg\left(t\right)t}{1+{tg}^2\left(t\right)}.(1+{tg}^2\left(t\right))dt$$$$=-2{\int }_0^{\frac{\pi }{4}}tg\left(t\right)tdtbypart=-2({[-tln\left(cost\right)]}_0^{\frac{\pi }{4}}$$$$+{\int }_0^{\frac{\pi }{4}}ln\left(cos\left(t\right)\right)dt$$$$=-\frac{\pi }{4}ln\left(2\right)-2{\int }_0^{\frac{\pi }{4}}ln\left(cos\left(t\right)\right)dt$$$$onewaytoofindit$$$$leta={\int }_0^{\frac{\pi }{4}}ln\left(cos\left(t\right)\right)dt$$$$={\int }_0^{\frac{\pi }{4}}\frac{1}{2}ln(sin\left(t\right)cos\left(t\right).\frac{1}{tg\left(t\right)})dt$$$$={\int }_0^{\frac{\pi }{4}}\frac{ln\left(\frac{sin\left(2t\right)}{2}\right)}{2}dt-\frac{1}{2}{\int }_0^{\frac{\pi }{4}}ln\left(tg\left(t\right)\right){dt}_{tg\left(t\right)=s}$$$$2t=u\Rightarrow \frac{1}{4}{\int }_0^{\frac{\pi }{2}}ln\left(u\right)du-{\int }_0^{\frac{\pi }{4}}ln\left(2\right)duinfirst$$$${\int }_0^{\frac{\pi }{4}}ln\left(tg\left(t\right)\right)dt={\int }_0^1\frac{ln\left(s\right)}{1+s^2}ds=\mathrm{\Sigma }{\int }_0^1{(-1)}^k{s}^{2k+1}ln\left(s\right)ds$$$$=-\mathrm{\Sigma }{(-1)}^k.\frac{1}{{(2k+1)}^2}=-\beta \left(2\right)=-G,Gcatalanconstante$$$$knowing{\int }_0^{\frac{\pi }{2}}ln\left(sin\left(t\right)\right)dtgiveuscloseforme$$$$2ndusefourierserieof$$$$ln\left(cos\left(t\right)\right)=-ln\left(2\right)-\sum _{n⩾1}\frac{cos\left(2(2n-1)t\right)}{2n-1}$$$$giveuseresulteimediatly$$$$withe\sum _{n⩾1}\frac{{(-1)}^{n-1}}{{(2n-1)}^2}=\beta \left(2\right)$$

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